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          <h1 class="post-title" itemprop="name headline">PAT A1010</h1>
        

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        <p>Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is “yes”, if 6 is a decimal number and 110 is a binary number.</p>
<p>Now for any pair of positive integers N1 and N2, your task is to find the radix of one number while that of the other is given.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. Each case occupies a line which contains 4 positive integers:<br>N1 N2 tag radix<br>Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set {0-9, a-z} where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number “radix” is the radix of N1 if “tag” is 1, or of N2 if “tag” is 2.</p>
<p>Output Specification:</p>
<p>For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print “Impossible”. If the solution is not unique, output the smallest possible radix.</p>
<p>Sample Input 1:<br>6 110 1 10<br>Sample Output 1:<br>2<br>Sample Input 2:<br>1 ab 1 2<br>Sample Output 2:<br>Impossible<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">//#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">#include &quot;iostream&quot;</span><br><span class="line">using namespace std;</span><br><span class="line">typedef long long LL;</span><br><span class="line">LL Map[256];</span><br><span class="line">LL inf = (1LL &lt;&lt; 63) - 1;//long long的最大值2^63-1</span><br><span class="line">void init()&#123;</span><br><span class="line">    for (char c = &apos;0&apos;; c &lt;= &apos;9&apos;; c++) &#123;</span><br><span class="line">        Map[c] = c - &apos;0&apos;;</span><br><span class="line">    &#125;</span><br><span class="line">    for (char c = &apos;a&apos;; c &lt;= &apos;z&apos;; c++) &#123;</span><br><span class="line">        Map[c] = c - &apos;a&apos; + 10;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line">LL convertNum10(char a[], LL radix, LL t)&#123;</span><br><span class="line">    LL ans = 0;</span><br><span class="line">    int len = strlen(a);</span><br><span class="line">    for (int i = 0; i &lt; len; i++) &#123;</span><br><span class="line">        ans = ans * radix + Map[a[i]]; //进制转换</span><br><span class="line">        if(ans &lt; 0 || ans &gt; t) return -1;//溢出或者超过N1的十进制</span><br><span class="line">    &#125;</span><br><span class="line">    return ans;</span><br><span class="line">&#125;</span><br><span class="line">int cmp(char N2[], LL radix,LL t)&#123;//N2的十进制与t比较</span><br><span class="line">    int len = strlen(N2);</span><br><span class="line">    LL num = convertNum10(N2, radix, t);//将N2转换为radix进制</span><br><span class="line">    if (num &lt; 0) return 1;//溢出，肯定是N2 &gt; t</span><br><span class="line">    if (t &gt; num) return  -1;//t较大，返回-1</span><br><span class="line">    else if(t == num) return 0;//相等，返回0</span><br><span class="line">    else return 1;//num较大，返回1</span><br><span class="line">&#125;</span><br><span class="line">LL binarySearch(char N2[], LL left, LL right, LL t)&#123;</span><br><span class="line">    LL mid;</span><br><span class="line">    while (left &lt;= right) &#123;</span><br><span class="line">        mid = (left + right)/2;</span><br><span class="line">        int flag = cmp(N2, mid, t);</span><br><span class="line">        if (flag == 0) return mid;</span><br><span class="line">        else if (flag == -1) left = mid + 1;</span><br><span class="line">        else right = mid - 1;</span><br><span class="line">    &#125;</span><br><span class="line">    return -1;//解不存在</span><br><span class="line">&#125;</span><br><span class="line">int findLargestDigit(char N2[])&#123;</span><br><span class="line">    int ans = -1, len = strlen(N2);</span><br><span class="line">    for (int i = 0; i &lt; len; i++) &#123;</span><br><span class="line">        if (Map[N2[i]] &gt; ans) &#123;</span><br><span class="line">            ans = Map[N2[i]];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    return ans+1;</span><br><span class="line">&#125;</span><br><span class="line">char N1[20], N2[20], temp[20];</span><br><span class="line">int tag,radix;</span><br><span class="line">int main()&#123;</span><br><span class="line">    init();</span><br><span class="line">    scanf(&quot;%s %s %d %d&quot;, N1, N2, &amp;tag, &amp;radix);</span><br><span class="line">    if (tag == 2) &#123;</span><br><span class="line">        strcpy(temp, N1);</span><br><span class="line">        strcpy(N1, N2);</span><br><span class="line">        strcpy(N2, temp);</span><br><span class="line">    &#125;</span><br><span class="line">    LL t = convertNum10(N1, radix, inf);//将N1从radix进制转换为十进制</span><br><span class="line">    LL low = findLargestDigit(N2);//找到N2中数位最大的位+1，当成二分下界</span><br><span class="line">    LL high = max(low, t) + 1;//上界</span><br><span class="line">    LL ans = binarySearch(N2, low, high, t);</span><br><span class="line">    if (ans == -1) printf(&quot;Impossible\n&quot;);</span><br><span class="line">    else printf(&quot;%lld\n&quot;,ans);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

      
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